MODULE X

MODULE X

CONTINUOUS SYSTEMS

In this module the equation of motion of continuous systems or distributed mass systems will be derived using both dAlembert principle and extended Hamiltons principles. Different one-dimensional systems such as longitudinal vibration of rod, transverse vibration of string, torsional vibration of rod and transverse vibration of Euler-Bernoulli beams will be considered in this module.

Lecture 1

Introduction to Continuous systems

In the previous modules we have studied about discrete mass system, which are modeled as single, two or multi-degrees of freedom systems. In these cases the system has a definite number of lumped masses, stiffness elements and damping elements. For example the cantilever beam with a tip mass as shown in Figure 10.1 is modeled as a single degree of freedom system with a spring and a mass. The stiffness k of the system was calculated using the following equation.

Here W is the weight of the attached mass, is the deflection of the beam with length L, Youngs modulus E and moment of inertia I. The natural frequency can be calculated using the formula where m is the attached mass. In this calculation we have neglected the mass of the beam. Hence it may be observed that by considering a point mass at the tip we obtained one natural frequency of the system. Instead of modeling this system as a single-spring mass if one consider the beam to be consist of several masses, then the system can be modeled as a multi-degree of freedom system as shown in figure 10.2(a). But as the dimension of each elemental mass considered in the above case is arbitrary, one may consider the beam as a continuous system with infinity number of distributed mass and stiffness and hence has infinity number of natural frequencies.

So in contrast to the discrete mass system, in distributed mass or continuous system the system has infinite number of natural frequencies and corresponding to each natural frequency, the system will have a distinct mode shape.

It may be observed that the response of the continuous system depends time and space coordinate (location). But in case of discrete system the response is only a function of time. Hence while the equation of motion of discrete systems are written in terms of ordinary differential equations, in case of continuous system they are written in terms of partial differential equations.

It may be noted that all the real systems are continuous system.

A continuous system for analysis purpose can be reduced to a finite number of discrete models. Each discrete model can be reduced to an eigen value problem.

In case of continuous system the solution yields infinite number of eigen values and eigen functions where as in discrete system the eigen values and eigen vectors are finite.

The concept of orthogonality is applicable to both discrete and continuous systems.

The eigen value problem in case of discrete system takes the from of algebraic equations while in continuous systems differential equations and some times integral equations are obtained. Eigenvectors of the discrete system becomes eigenfunction of the continuous system.

The response of the system will depends on the boundary conditions. There are two different types of boundary conditions viz., geometric boundary conditions and natural boundary conditions. Geometry boundary conditions also known as essential or imposed boundary conditions result from conditions of purely geometric compatibility. For example in case of a clamped-clamped beam in both the ends deflections and slopes are zero (Fig. 10.3).

Fixed-Fixed beam : Deflections and slope at both ends are zero which constitute geometric boundary conditions

Free-free beam Bending moment and shear forces are zero at both ends (natural boundary conditions at both ends)

Simply supported beam

(mixed boundary conditions)

Fig. 10.3

Natural boundary condition also known as additional or dynamic boundary condition, which results from the balance of moments or forces in the boundary. For example in case of a free-free beam which may be a model of a flying aeroplane or a spacecraft, at both the ends in this system shear force and bending moments are zero (Fig. 10.3). Hence they constitute the natural boundary conditions.

In some systems one may find both the geometric and natural boundary conditions. For example in case of a simply supported beam, both deflection and bending moment at end points are zero. Hence the boundary condition in this case may be termed as mixed boundary conditions. Now let us consider specific examples of continuous one-dimensional systems.

Lateral vibration of a flexible taut string

T T

L

Fig. 10.4

Strings are mostly used in musical instruments and many other applications of domestic and industrial in nature. A string of length L is shown in Figure 10.4(a), which is subjected to tension T. Let at time t = 0, the string is pulled in the lateral direction (y direction) as shown in figure 10.4 (b) and left. Hence the lateral deflection u along the string is a function of the space variable x and time t i.e.,

If the lateral deflection is small, the change in tension T due to the deflection is negligible. Figure 10.4(c) shows the free body diagram of an elemental length of the string. When the string is vibrating, in the y direction inertia force is acting. Considering the forces in the vertical direction, applying Newtons second law one may have

(10.1)

Here m is the mass per unit length of the string. Now assuring small deflections u and slope (, the equation 10.1 reduces to following equation.

(10.2)

Now substituting the slope in equation 10.2 one may write

or,

or,

or,

The above equation is known as Wave equation. One may use Hamiltons principle to derive the equation of motion of this system also.

Longitudinal vibration of rod

Consider a rod of length l (Fig. 10.5) subjected to a force F at time t=0 and then released. It will be subjected to longitudinal vibration.

x dx

F

Fig. 10.5 Longitudinal vibration of rod

Let u(x,t) be the axial displacement of an element dx of the rod.

From Hooks law (1)

Applying Newtons second law (Fig. 10.6)

(2)

(3)

P

u

Fig. 10.6 Displacement of dx

P= force at x

A= Cross sectional area

E= Youngs Modulus.

d is the mass of the element

= Mass per unit volume.

From equation (3) If AE is const

(4)

or (5)

or where (6)

It may be observed that we are getting the same wave equation in this case where only c is different. It can be shown that c represent the velocity of the wave in the rod.

Lecture M10 L2

Derivation of equation of motion by Hamiltons Principle

Use Hamiltons Principle to find equation of motion of a rod in longitudinal vibration.

Sol.:

Kinetic Energy T =

=..(7)

P.E. = strain energy stored

EMBED Equation.DSMT4

EMBED Equation.DSMT4 (8)

(9)

(10)

where , A, E are assumed constants.

Using Extended Hamiltons Principle

(11)

where

or, (12)

or, (13)

Asssuming EA to be constant, as is arbitrary , setting the coefficient of equal to zero will yield the equation of motion.

or, where (14)

It may be noted from equation (13) that along with the equation of motion, one may get a set of boundary conditions. The boundary conditions for some of the typical cases are given in the following Table.

Table 1 Boundary Conditions for the longitudinal vibration of rod

CaseBoundary condition left

x=0Boundary condition right

x=l

Free end

Fixed end

End spring

K

End man

End dampes

Torsional vibration of shafts

let ((x,t) be the angular displacement of an element dx of the shaft.

(15)

The twist of an element length

(16)

Here is the torsional stiffness of the shaft, is shear modulus of the material and is the polar moment of inertia of the rod.

Applying Newtons second law for the rotation in the ( direction

(17)

Here = mass per unit volume

(18)

or (19)

or where (20)

So the equation of motion of all the three cases discussed can be expressed as

( is a linear differential operator on function .

The boundary conditions in case of torsional vibration of rod are given in Table 2.

Table 2: Boundary conditions for Torsional Vibration of rods

CaseBoundary condition left

x=0Boundary condition right

x=l

Fixed end

Free end

Torsional spring

k

Inertia Jp

Jp

Torsional dampoo

Lecture M10L3

Solution of Wave EquationIn the last two lectures the equation of motion of lateral vibration of string, longitudinal vibration of rod and torsional vibration of shafts were carried out using Newtons law and Hamiltons principle. In all these three cases the equation of motion of the system reduces to that of Wave equation, which can be given by

(1)

where the values of C2 are for different cases are given in the following Table.

Case C2

Lateral Vibration of taut String (T/)

Longitudinal Vibration of rodE/

Torsional Vibration of RodG/

To find the response of different system one may use the variable separation method by using the following equation.

(2)

is known as the mode shape of the system andis known as the time modulation. Now equation (1) reduces to

(3)

or (4)

Since the left side of equation (4) is independent of time t and the right side is independent of x the equality holds for all values of t and x. Hence each side must be a constant. As the right side term equals to a constant, it implies that the acceleration term is proportional to displacement term , one may take the proportionality constant equal to to have a simple harmonic motion in the system. If one takes a positive constant, the response will grow exponentially and make the system unstable. Hence one may write equation (4) as

(5)

Hence, (6)

And (7)

The solution of equation (6) and (7) can be written as

(8)

(9)

Hence,

. (10)

The coefficients and are to be determined from the boundary conditions and the coefficients and are to be evaluated from the initial conditions. Now let us consider several examples to study the free vibration response of strings, longitudinal vibration of rods and torsional vibration of shaft.

Example 1: Find the mode shape of a string fixed at both the ends.

Solution:

For the string fixed at both the ends the boundary conditions are displacement at both the ends are zero.

Hence, and . ()

This implies that

and . ( a,b )

Now from equation (9)

(c)

Substituting condition (a) in equation (c) one obtains . (d)

Now substituting (b) and (d) in (c) yields

(e)

Here constant is not equal to zero as this will lead to the trivial solution of the system. Hence

(f)

or, (g)

So, (13)

And the mode shape is given by

(14)

Hence the string will have a number of frequencies. The fundamental frequency can be given by and the overtones are integer multiples of the fundamental frequency. So it may be noted that the frequency of the string depends on three factors viz., tension in the string, mass per unit length of the string material and length of the string. In the musical instrument as is constant and is kept constant, the artist vary the length of the string to produce a mixture of different overtones.

Equation (14) represents the eigen modes of the fixed-fixed string. Hence one may write the frequency equation as

(15)

and eigen function for a string fixed at both ends as

(16)

It may be noted that in equation 15, the constants are arbitrary. Like in the case of multi degree of freedom system, here also one may normalize the eigen modes by using the following expression

(17)

It may be recalled that in case of weighted modal matrix . Similarly to normalize the eigen function one can use equation (17).

So in this case now substituting equation (16) in equation (17), one obtains

(18)

(19)

(21)

Equation (19) can be written as

(22)

or, (23)

or, (24)

or, (25)

Hence the normalized modes of the lateral vibration of string are

(26)

Orthogonal properties of the eigen modes

As it is observed in the multi degree of freedom system that the eigenmodes are orthogonal here also it can be shown that the eigenfunctions are orthogonal.

When r=s

=1

Hence we can write

r,s=1

is the Kronecker delta

To verify equation

Where

This shows the orthogonality property of the eigen modes.

Lateral Vibration of Euler-Bernoulli Beam

Lateral vibration of beam was discussed by Euler in 1744 and Daniel Bernoull in 1751 Eqn( below appears to have been introduced by Jacob Bernoullin 1789.

y

u

x

xdx

When the deflection of the beam is assumed to be due to bending, then the beam is called an Euler Bernoulli beam.

From strength of material it is known

In the Y direction net force Fy

Applying Newtons second law

m=mass per unit length of the beam

,

or,

Exercise problem: Derive the Equation of motion from Hamiltons Principle

Hints Kinetic energy

Potential energy

Essentially the problem of the vibration of a one-dimensional structural member can be described by a linear hyperbolic partial differential equation

Boundary conditions on the geometric configuration (displacement or slope) are Dirichlet or geometric boundary conditions.

Boundary conditions on the forces and moments are Neumann or Natural boundary conditions.

Combinations of the two types are mixed boundary conditions.

The boundary conditions for transverse vibration is given in Table 3.

Table 3 Boundary conditions for transverse vibration of beam

Case BCs left x=0 BC.s right x=l

Clamped (deflection, slope=0)

Pinned (deflection, moment=0

Sliding (slope, Shear =0)

Free (moment, shear =0)

Mass m and moment of inertia Jp

m,Jp

Damper c and spring k

Lecture M10L4

Transverse vibration of Beam with various boundary conditions

Solution of Euler Bernoulli equation

In the last two lectures the equation of motion of lateral vibration of string, longitudinal vibration of rod and torsional vibration of shafts were carried out using Newtons law and Hamiltons principle. In all these three cases the equation of motion of the system reduces to that of Wave equation, which can be given by

(1)

To find the response of the system one may use the variable separation method by using the following equation.

(2)

is known as the mode shape of the system andis known as the time modulation. Now equation (1) reduces to

(3)

or (4)

Since the left side of equation (4) is independent of time t and the right side is independent of x the equality holds for all values of t and x. Hence each side must be a constant. As the right side term equals to a constant implies that the acceleration is proportional to displacement , one may take the proportionality constant equal to to have simple harmonic motion in the system. If one take a positive constant, the response will grow exponentially and make the system unstable. Hence one may write equation (4) as

(5)

Hence, (6)

And (7)

Taking (8)

The above equation can be written as

(9)

The solution of equation (6) and (9) can be given by

(10)

(11)

Hence,

. (12)

Here constants and can be obtained from the initial conditions and constantscan be obtained from the boundary conditions. Let us now determine the mode shapes of simply supported beam, fixed-fixed beam, cantilever beam and free-free beams.

Simply supported beam

For beam simply supported at both the end the boundary conditions can be given by

13 (a-d)

Using equation 2 in 13(a-d) the boundary conditions reduces to

14(a-d)

Now using the expression for the mode shape (equation 11) and its second order derivative can be given by the following expression.

. (15)

From 14(a)

From 14(b)

Hence both (16)

Now from 14(c) and 14(d)

17(a,b)

From equation 17, (18)

As shown in Figure 1It may be noted that the hyperbolic function is not equal to zero. Hence .

Now as out of the four constants three constants A, B and D are zero, the other remaining constant C should not be equal to zero. It may be noted that C = 0 correspond to the trivial solution i.e, u = 0 of the system. As we are studying the vibration of the system i.e, about the nontrivial solution of the system, from equation (18) one may obtain

(19)

Hence, (20)

Figure 2

Now from equation 8 and 20 one may write the expression for the frequency as

(21)

(22)

and the mode shape can be given by

(23)

Hence from equation (22) and (23) it may be noted that the simply supported system has a large number of frequency and corresponding mode shapes. As are the system parameters and are fixed for a particular system, the frequency of the nth mode is n2times the fundamental frequency. For example, the second mode frequency is 4 times the fundamental or first mode frequency and the third mode frequency is 9 times the fundamental frequency.

Fig. First four modes of simply supported beam

Cantilever beam

In case of cantilever beam the boundary conditions are

At left end i.e.,

(24)

At the free end i.e.,

(25)

(26)

(27)

Substituting these boundary conditions in the general soulution

(28)

From (26) (29)

(30)

From (27 and 30) one may have

(31)

or, (32)

Hence one may solve the frequency equation to obtain frequencies of different modes. For the first five modes the values of are calculated as 1.875, 4.694, 7.855, 10.996, 14.137.

Since for higher roots will be quite high, so . Hence for more than 5th mode one may write

for n>5 (33)

In this case the first 4 modes are shown in the following figures. The points with zero displacements i.e., the node points are marked by circles.

Here the expression for mode shape can be given by

(34)

which are plotted in the above figure.

Fixed at both the ends

Applying the fixed-fixed boundary condition and proceeding in the way mentioned in the above two boundary conditions one will get the frequency equation as

(35)

The first five roots ofare

= 0, 4.7300, 7.853, 10.996, 14.137, 4.730

For higher values for n>5

The first solution =0 represents the rigid body mode of the fixed-fixed beam.

Thus the higher modes of a fixed-fixed beam have approximately the same frequencies as those of the cantilever beam with same physical properties. But the mode shapes are different.

The mode shapes are shown below.

Exercise Problems: In case of fixed-fixed beam the mode shape expression can be given by

(36)Free-Free beam

Frequencies for free-free beam are same as those of fixed-fixed beam, but the displacement curves are different as the boundary conditions are different.

Exercise problem: Show that the frequency equation in case of a free-free beam can be given by

The first five roots of this equation are

4.7300, 7.853, 10.996, 14.137, 4.730

For higher values for n>5Summary

Rigorously speaking, all real system are continuous system

A continuous system for analysis purpose can be reduces to a finite number of discrete models. Each discrete model can be reduced to an eigenvalue problem.

In case of continuous systems the solution yields infinite number of eigenvalues and eigenfunctions where as in discrete system the number of eigenvalues and eigenvectors are finite.

The concept of orthogonality is applicable to both discrete and continuous systems.

The eigenvalue problem in case of discrete system takes the from of algebraic equations while in continuous systems differential equations and some times integral equations are obtained. Eigenvectors of the discrete system becomes eigenfunction of the continuous system.

For one dimensional systems like transverse vibration of taut string, longitudinal vibration of rod, torsional vibration of rod the equation of motion reduces to the Wave equation i.e.,

EMBED Equation.3 where C are given in the following table. Case C2

L()

Lateral V. of taut String (T/)

Longitudinal V. of rodE/

Lateral Vib. BeamEI/

Torsconal Vib. RodG/

Exercise Problems

1. A shaft 40 mm diameter and 2.5 m long has a mass of 15 kg per m length. It is fixed at both the ends and carries three masses 90 kg, 140 kg and 60 kg at 0.8 m, 1.5 m and 2 m respectively from the left support. Taking E = 200 GN/m2, find the frequency of the transverse vibrations. (Hinds: (L = 4.730).

2. A rotor has a mass of 12 kg and is mounted midway on a 24 mm diameter horizontal shaft supported at the ends by two bearings. The mass of the shaft is 2 kg and bearings are 1 m apart. The shaft rotates at 2400 rpm. If the center of mass of the rotor is 0.1 mm away from the geometric center of the rotor due to a certain manufacturing defect, find the amplitude of steady state vibration and the dynamic force transmitted to the bearing. Take ( = 0.01 and E = 200 GN/ m.

3. A rotor has a mass of 15 kg and is mounted midway on a 24 mm diameter horizontal shaft supported at the ends by two bearings. The mass of the shaft is 2 kg and bearings are 1 m apart. Find the first two natural frequency using energy principle. E = 200 GN/ m.

4. Derive the equation of motion of the following system using extended Hamiltons principle (a) Lateral vibration of taut string (b) torsional vibration of shaft (c) transverse vibration of a cantilever beam with tip mass and supported by spring in the middle.

5. Find the normalized mode shapes in case of (a) longitudinal vibration of rod (b) transverse vibration of beam with (i) simply supported boundary conditions, (ii) fixed-fixed boundary conditions, (iii) fixed-free end conditions (iv) free-free end conditions.

Fig 10.1

Fig. 10.2(a,b)

EMBED Equation.3

y

x

(a)

EMBED Equation.3

T

T

(

EMBED Equation.3

Fig. 10.7

dx

x

T

(

EMBED Equation.3

(

l

dx

Please modify the spring

EMBED Equation.DSMT4

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

EMBED Equation.3

Figure 1

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